Tính tích phân sau: ∫(sinx.sin3x/3^x+1)dx

Question

Tính tích phân sau:

\( \int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\frac{\sin x.\sin 3x}{3^{x} +1}  dx \)

Answer 1

\(I=\int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\frac{\sin x.\sin 3x}{3^{x} +1}  dx\)

\(=\frac{1}{2} \left(\int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\frac{\cos 2x}{3^{x} +1}  dx-\int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\frac{\cos 4x}{3^{x} +1}  dx\right) \)
Xét \(I_{1}^{} =\int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\frac{\cos 2x}{3^{x} +1}  dx. \)Đặt \(t=-x\Rightarrow dx=-dt. \)

Ta có: \(  I_{1}^{} =\int _{-\frac{\pi }{2} }^{0}\frac{\cos 2x}{3^{x} +1}  dx+\int _{0}^{\frac{\pi }{2} }\frac{\cos 2x}{3^{x} +1}  dx  (*). \)

\(\Rightarrow I_{1a}^{} =\int _{-\frac{\pi }{2} }^{0}\frac{\cos 2x}{3^{x} +1}  dx=-\int _{\frac{\pi }{2} }^{0}\frac{\cos 2t}{3^{-t} +1}  dt\)

\(=\int _{0}^{\frac{\pi }{2} }\frac{3^{t} .\cos 2t}{3^{t} +1}  dt=\int _{0}^{\frac{\pi }{2} }\frac{3^{x} .\cos 2x}{3^{x} +1}  dx. \)
Thay vào (*) ta được:

\(I_{1}^{} =\int _{0}^{\frac{\pi }{2} }\frac{3^{x} .\cos 2x}{3^{x} +1}  dx+\int _{0}^{\frac{\pi }{2} }\frac{\cos 2x}{3^{x} +1}  dx\)

\(=\int _{0}^{\frac{\pi }{2} }\cos 2x dx=\frac{1}{2} \sin 2x\left|\begin{array}{l} {\frac{\pi }{2} } \\ {0} \end{array}\right. =0.\)

 Xét \(I_{2}^{} =\int _{-\frac{\pi }{2} }^{\frac{\pi }{2} }\frac{\cos 4x}{3^{x} +1}  dx.\)

Đặt \( t=-x\Rightarrow dx=-dt.\)

\( \Rightarrow I_{2}^{} =\int _{-\frac{\pi }{2} }^{0}\frac{\cos 4x}{3^{x} +1}  dx+\int _{0}^{\frac{\pi }{2} }\frac{\cos 4x}{3^{x} +1}  dx\)

\(=\int _{0}^{\frac{\pi }{2} }\cos 4x. dx=\frac{1}{4} \sin 4x\left|\begin{array}{l} {\frac{\pi }{2} } \\ {0} \end{array}\right. =0\)

 Vậy \(I=0.\)